Estimation Of Materials Required For LT Lines Or Low Tension Lines

Hello everyone, in this article, I will discuss the estimation of materials required for LT lines and also methods of calculating. You can also catch me @ Instagram – Chetan Shidling.

Before starting the article if you want to watch a video related to electrical estimation and costing, you can watch my YouTube videos – Click Here

Also read- Electrical Machines Questions And Answers.

Estimation Of Materials Required For LT Lines

1. No of spans
= Length of line ÷ span

What is the span?
Span means horizontal distance between two poles is called span. A number of span means for example in 1000 KM line how many spans is there.

2. No of supports

= No of spans + 1 ( for taping pole )

Supports means nothing but poles which give support to the conductors. No support means in 1000 KM line how many supports are there.

3. R.C.C poles are used at starting, deviation points and anchoring points
R.C.C full form is reinforced cement concrete, this pole is constructed using concrete and with some rods. This pole very strong compared to the other pole. R.C.C pole cannot break easily, it can hold heavyweight.

4. P.C.C poles are used at intermediate poles only.

P.C.C pole full form is plain cement concrete, this pole is weak in tension and it is constructed only by using cement concrete.

5. No of cross-arms

No of 4-pin cross-arm = No of support + 1 ( tapping pole )
cross-arms which gives support to the insulators and conductors. cross arms are classified into many types; V shape cross arm, horizontal cross arm, U-shaped cross arm, etc.

6. No of pin insulators

= 4 [Number of supports + no of deviations – (one dead-end + anchoring points)]

Pin insulators that do not allow an electric current to flow through it, pin insulators are used if the voltage is below 33 KV. 

7. No of strain insulators

= [Two (one for each dead end) + 2(number of anchoring points )]8. Length of the conductors required :

(a). For phases, the ACSR No.2 conductor has required = 3 × lengths of line(b). For neutral, ACSR No.4 conductor has required = 1 × length of a line
The full form of ACSR conductor is “Aluminium Conductor Steel Reinforced”. 

9. The number of guy sets

= 2(one for each dead end) + (Anchoring point)2 + deviation points.

10. Whenever guarding is to be provided two numbers of special cross arms with two numbers of guy sets and 15 kg 8 SWG G.I wire are to be used.
A guarding is provided for the safety of life, installation, and communication circuits.

The guarding for 11 KV lines is provided at road crossings, canal crossings, railway crossing, crossing over lt lines, or communication lines.

This was about materials required for LT lines.

Problem On Estimation Of Materials Required For LT Lines

Q: A three phases 4 wire 415/230 V LT line has to be extended for a length of 1 KM, for electrification of a factory. Assuming an average span of 60 meters. Prepare the list of materials required for the same.

Problem On Estimation Of Materials Required For LT Lines

Solution :

1. Number of spans 

= Length of line / span

= (1 × 1000) / 60 

16.66 say 17 No’s

( span means distance between two poles)  

The number of spans in the one-kilometer line is 17 No.

2. Total number of supports 

= Number of spans + 1 

= 17 + 1 = 18 No’s

The total number of supports required for a one-kilometer line is 18 No’s.

3. Number of R.C.C poles 

= one for tapping + one for dead end + ( at 0.5 KM anchoring ) = 1 + 1 + 1 = 3 No’s

The total number of reinforced cement concrete pole required is 3 No’s.

4. Number of P.C.C poles 

= Number of supports – R.C.C poles

= 18 – 3 = 15 No’s

The number of plain cement concrete pole required is 15 No’s.

5. Number of 4-pin cross arms = 18 No’s

How many poles are there that many cross arms are required.

6. Number of 1.1 KV class guy sets

= 1(starting pole) + 1(terminating pole) + 2(at 0.5 KM anchoring pole) = 4 No’s

The number of guy set required is 4 No’s.

7. Number of pin insulators

= (No of intermediate poles + Anchoring ) × 4

= (15 + 1) × 4

= 16 × 4 = 64 No’s

The number of pin insulator required is 64 No’s

8. Number of strain insulators

= 4 * [starting pole + end pole + 2(Anchoring pole)] = 4(2 + 2) = 16 No’s

9. Length of conductors

(a). A.C.S.R No.2 weasel conductors for 3-phase conductors with 3% sag = 3 × 1 KM × 1.03 = 3.09 KM

(b). A.C.S.R No.4 squirrel conductor for neutral with 3% sag  = 1 × 1 KM × 1.03 = 1.03 KM

10. Miscellaneous materials such as nuts, bolts, barbed wires, etc.

List Of Materials Required For LT Lines

SI.NoList Of MaterialsUnitQuantity
01R.C.C poles 7.5 meter long, 140 kg working loadNo’s03
02P.C.C poles 7.5 meters long, 115 kg working loadNo’s15
034-pin cross arms with pole clamp nut, bolt, and washersSet18
041.1 KV class guy set completeSet04
051.1 KV pin insulators with pin and nuts complete setSet64
06No 8 strain insulator of porcelainNo’s16
07No 2 A.C.S.R squirrel conductor with 3% sag for all line conductorKM3.09
08No A.C.S.R squirrel conductor with 3% sag for the neutral conductorKM1.03
09Miscellaneous materials such as nut, bolts, washers, barbed wire, etcLump sum45

I hope this article may help you all a lot…….. Thank you for reading. If you have any doubts related to this article “estimation of materials required for LT lines”, then comment below.

Tags: Estimation Of Materials Required For LT Lines.

Also, read:

Chetan Shidling

Learning Addicted, Engineer, Blogger, YouTuber, Content Writer, Google Adsense Expert, SEO Expert, Digital Marketer, Web Developer, App Developer, Cloud Computing, and a lot more...

Leave a Reply

Your email address will not be published. Required fields are marked *